Setting Up Correct Equation...2

[mathdad]

For questions 64, 68, and 70, set up the correct equations. Use any variable(s) of choice.

64. A recently retired couple needs 12,000 per year to supplement their Social Security. They have 150,000 to invest to obtain this income. They have decided on two investment options: AA bonds yielding 10 percent per annum and a Bank Certificate yielding 5 percent.

(a) How much should be invested in each to realize exactly 12,000?

(b) If, after 2 years, the couple requires 14,000 per year in income, how should they reallocate their investment to their investment to achieve the new amount?

Part A Set Up:

x + y = 12,000
0.10x + 0.05y = 150,000

Part B Set Up:

x + y = 14,000
0.10x + 0.05y = 150,000

68. One group of people purchased 10 hot dogs and 5 soft drinks at a cost of 35.00. A second bought 7 hot dogs and 4 soft drinks at a cost of 25.25. What is the cost of a single hot dog? A single soft drink?

Set Up:

Let h = hot dogs

Let s = soft drinks

10h + 5s = 35.00
7h + 4s = 25.25

70. Pamela requires 3 hours to
swim 15 miles downstream on the Illinois River. The return trip upstream takes 5 hours. Find Pamela’s average speed in still water. How fast is the current? (Assume that Pamela’s
speed is the same in each direction.)

Set Up:

Let a = average speed in still water

Let c = speed of current

3(a + c) = 15
5(a - c) = 15

Is any of this correct?

 

[JeffM]

In part A, your set up makes no sense. You have not defined your symbols, a potential recipe for confusion. Based on your first equation, x + y = 12,000, you seem to have decided that x and y are the annual incomes from the two investments although there is no clue as to which investment each is associated with.

Nevertheless, under that labeling, it should be obvious that neither x nor y can be negative and neither can exceed 12,000. How can you expect to multiply two numbers less than 12,000 by numbers far less than 1 and have them add to 150,000. Thinking about what your equations mean will not assure that they are correct, but such reasonableness checks will identify when you are really off track.

You can have your basic variables be incomes, but that is the hard way to do it.

AND PLEASE ONE PROBLEM PER THREAD

 

[Dr.Peterson]

68. One group of people purchased 10 hot dogs and 5 soft drinks at a cost of 35.00. A second bought 7 hot dogs and 4 soft drinks at a cost of 25.25. What is the cost of a single hot dog? A single soft drink?

Set Up:
Let h = hot dogs
Let s = soft drinks
10h + 5s = 35.00
7h + 4s = 25.25

Again, the definitions are not clear, which could easily lead to confusion. I'd write (even if only for my own use):

Let h = cost of one hot dog​

Let s = cost of one soft drink​

Your use of plurals strongly suggests that you meant quantities!

But your equations are correct, if you meant what I said.

70. Pamela requires 3 hours to swim 15 miles downstream on the Illinois River. The return trip upstream takes 5 hours. Find Pamela’s average speed in still water. How fast is the current? (Assume that Pamela’s speed is the same in each direction.)

Set Up:
Let a = average speed in still water
Let c = speed of current
3(a + c) = 15
5(a - c) = 15

I'd probably want to include the units, mph, in the definitions, especially since nothing otherwise states that unit, and you could easily forget it in writing your answer.

But again, the equations are good. (And in solving, I would not distribute, but would divide by the 3 and the 5, as many students don't think to do.)

 

[mathdad]

In part A, your set up makes no sense. You have not defined your symbols, a potential recipe for confusion. Based on your first equation, x + y = 12,000, you seem to have decided that x and y are the annual incomes from the two investments although there is no clue as to which investment each is associated with.

Nevertheless, under that labeling, it should be obvious that neither x nor y can be negative and neither can exceed 12,000. How can you expect to multiply two numbers less than 12,000 by numbers far less than 1 and have them add to 150,000. Thinking about what your equations mean will not assure that they are correct, but such reasonableness checks will identify when you are really off track.

You can have your basic variables be incomes, but that is the hard way to do it.

AND PLEASE ONE PROBLEM PER THREAD

Problem 64 threw me into a loop with so many words and two parts. I get lost when there are too many words mixed in with more than 3 numbers. To be honest, I had no idea what I was doing for 64.

 

[mathdad]

Again, the definitions are not clear, which could easily lead to confusion. I'd write (even if only for my own use):

Let h = cost of one hot dog​

Let s = cost of one soft drink​

Your use of plurals strongly suggests that you meant quantities!

But your equations are correct, if you meant what I said.


I'd probably want to include the units, mph, in the definitions, especially since nothing otherwise states that unit, and you could easily forget it in writing your answer.

But again, the equations are good. (And in solving, I would not distribute, but would divide by the 3 and the 5, as many students don't think to do.)

What about 64? Is my set of equations correct for 64? I get lost when word problems are too lengthy (involve too words and numbers in the mix). What is the best way to tackle lengthy questions like 64?

 

[Dr.Peterson]

What about 64? Is my set of equations correct for 64? I get lost when word problems are too lengthy (involve too words and numbers in the mix). What is the best way to tackle lengthy questions like 64?

JeffM told you what was wrong with 64. Define your variables, and state what each equation means, before writing it. You should take x and y to be the amounts invested.

The way to tackle a long question is to take one part at a time.

 

[mathdad]

JeffM told you what was wrong with 64. Define your variables, and state what each equation means, before writing it. You should take x and y to be the amounts invested.

The way to tackle a long question is to take one part at a time.

I'll work on this later.

 

[JeffM]

@mathdad
When I tutor face to face, I teach a non-traditional way to deal with word problems. It is non-traditional because it requires knowing how to deal with systems of equations, which is usually deferred for quite some time. But you obviously know how to deal with systems of equations.

Step 1: Identification and Labeling. Think about the problem and identify the relevant quantities that are unknown. For each such quantity, assign a letter in writing. It is now far harder to get confused about meaning because you have a name for everything. Normal human thought is done using names. So in this problem I would start

x = income from bonds
y = income from certificate

p = amount invested in bonds
q = amount invested in certificate

I have not tried yet to do a lick of math, but I am now clear in what is unknown and can think about it in simple terms.

Step 2: Translation. In a word problem, we are given information in a natural language. To use mathematics, we must translate that information into the language of mathematics. We have four unknowns. To use algebra, we need four equations.

p + q = 150000
x + y = 12000
x = 0.1p
y = 0.05q

Step 3: Solution. Now we do algebra. Can you solve that system of equations?

Step 4: Verification. Check that the answers satisfy the equations in step 2.

Every single word problem, no matter how many unknowns it has, can be solved in this way.

 

[mathdad]

@mathdad
When I tutor face to face, I teach a non-traditional way to deal with word problems. It is non-traditional because it requires knowing how to deal with systems of equations, which is usually deferred for quite some time. But you obviously know how to deal with systems of equations.

Step 1: Identification and Labeling. Think about the problem and identify the relevant quantities that are unknown. For each such quantity, assign a letter in writing. It is now far harder to get confused about meaning because you have a name for everything. Normal human thought is done using names. So in this problem I would start

x = income from bonds
y = income from certificate

p = amount invested in bonds
q = amount invested in certificate

I have not tried yet to do a lick of math, but I am now clear in what is unknown and can think about it in simple terms.

Step 2: Translation. In a word problem, we are given information in a natural language. To use mathematics, we must translate that information into the language of mathematics. We have four unknowns. To use algebra, we need four equations.

p + q = 150000
x + y = 12000
x = 0.1p
y = 0.05q

Step 3: Solution. Now we do algebra. Can you solve that system of equations?

Step 4: Verification. Check that the answers satisfy the equations in step 2.

Every single word problem, no matter how many unknowns it has, can be solved in this way.

I would substitute the value for x and y into x + y = 12,000.

p + q = 150000..............A
0.1p + 0.05q = 12000....B
x = 0.1p..........................C
y = 0.05q........................D

I would begin by solving equations A and B for p and q. I then can plug the value of p and q into
x = 0.1p and y = 0.05q to find x and y.

 

[JeffM]

I would substitute the value for x and y into x + y = 12,000.

p + q = 150000..............A
0.1p + 0.05q = 12000....B
x = 0.1p..........................C
y = 0.05q........................D

I would begin by solving equations A and B for p and q. I then can plug the value of p and q into
x = 0.1p and y = 0.05q to find x and y.

I was hoping that you would at this point stop using words and use algebra. That was the whole point of labeling and translating. We replace the virtually mechanical process of manipulating symbols for the difficult process of thinking systematically in terms of words. The sentence about substituting the values of x and y into x + y = 12000 makes no sense because you do not know those values yet.

But your final paragraph probably indicates that you do know what to do. You systematically reduce the number of unknowns by finding out from any one of the equations what one unknown means in terms of one or more of the other unknowns and then replacing that unknown in all the remaining equations. We keep doing this until we are down to one equation and then solve that numerically.

[MATH]y = .05q \implies x + 0.05q = 12000, \ p + q = 150,000, \text { and } x = 0.1p[/MATH]
We do that process again.

[MATH]x = 0.1p \implies 0.1p + 0.05q = 12000 \text { and } p + q = 150000[/MATH]
Now what?

 

[mathdad]

I was hoping that you would at this point stop using words and use algebra. That was the whole point of labeling and translating. We replace the virtually mechanical process of manipulating symbols for the difficult process of thinking systematically in terms of words. The sentence about substituting the values of x and y into x + y = 12000 makes no sense because you do not know those values yet.

But your final paragraph probably indicates that you do know what to do. You systematically reduce the number of unknowns by finding out from any one of the equations what one unknown means in terms of one or more of the other unknowns and then replacing that unknown in all the remaining equations. We keep doing this until we are down to one equation and then solve that numerically.

[MATH]y = .05q \implies x + 0.05q = 12000, \ p + q = 150,000, \text { and } x = 0.1p[/MATH]
We do that process again.

[MATH]x = 0.1p \implies 0.1p + 0.05q = 12000 \text { and } p + q = 150000[/MATH]
Now what?

Part A

p + q = 150,000
x + y = 12 000
x = 0.1p
y = 0.05q

p + q = 150,000
0.1p + 0.05q = 12 000
x = 0.1p
y = 0.05q

p = 150,000 - q

0.1(150,000 - q) + 0.05q = 12,000

Solving for q, I got 60,000.

I plug 60,000 for q in either equation above.

p + q = 150,000

p + 60,000 = 150,000

p = 150,000 - 60,000

p = 90,000

Part B

p + q = 150,000
x + y = 14, 000
x = 0.1p
y = 0.05q

x + y = 14,000

0.1p + 0.05q = 14,000

p + q = 150,000

p = 150,000 - q

0.1(150,000 - q) + 0.05q = 14,000

Solving for q, I got 20,000.

p + q = 150,000

p + 20,000 = 150,000

p = 150,000 - 20,000

p = 130,000

 

[JeffM]

Part A

p + q = 150,000
x + y = 12 000
x = 0.1p
y = 0.05q

p + q = 150,000
0.1p + 0.05q = 12 000
x = 0.1p
y = 0.05q

p = 150,000 - q

0.1(150,000 - q) + 0.05q = 12,000

Solving for q, I got 60,000.

I plug 60,000 for q in either equation above.

p + q = 150,000

p + 60,000 = 150,000

p = 150,000 - 60,000

p = 90,000

Part B

p + q = 150,000
x + y = 14, 000
x = 0.1p
y = 0.05q

x + y = 14,000

0.1p + 0.05q = 14,000

p + q = 150,000

p = 150,000 - q

0.1(150,000 - q) + 0.05q = 14,000

Solving for q, I got 20,000.

p + q = 150,000

p + 20,000 = 150,000

p = 150,000 - 20,000

p = 130,000

You forgot my fourth step: verification.

[MATH]p = 90000 \text { and } q = 60000 \implies p + q = 90000 + 60000 = 150000. \ \checkmark[/MATH]
[MATH]0.1p + 0.05q = 0.1 * 90000 + 0.05 * 60000 = 9000 + 3000 = 12000. \ \checkmark[/MATH]
And for part b

[MATH]p = 130000 \text { and } q = 20000 \implies p + q = 1300000 + 20000 = 150000. \ \checkmark[/MATH]
[MATH]0.1p + 0.05q = 0.1 * 130000 + 0.05 * 20000 = 13000 + 1000 = 14000. \ \checkmark[/MATH]
Well done.

Isn't that process starightforward?

 

[mathdad]

You forgot my fourth step: verification.

[MATH]p = 90000 \text { and } q = 60000 \implies p + q = 90000 + 60000 = 150000. \ \checkmark[/MATH]
[MATH]0.1p + 0.05q = 0.1 * 90000 + 0.05 * 60000 = 9000 + 3000 = 12000. \ \checkmark[/MATH]
And for part b

[MATH]p = 130000 \text { and } q = 20000 \implies p + q = 1300000 + 20000 = 150000. \ \checkmark[/MATH]
[MATH]0.1p + 0.05q = 0.1 * 130000 + 0.05 * 20000 = 13000 + 1000 = 14000. \ \checkmark[/MATH]
Well done.

Isn't that process starightforward?

Yes, I forgot step 4. I had no intention to solve any of the new posted threads. My main concern is the set up. If the set up is right, then I will find the right answer(s) via algebra.

 

[JeffM]

Yes, I forgot step 4. I had no intention to solve any of the new posted threads. My main concern is the set up. If the set up is right, then I will find the right answer(s) via algebra.

Yes. So what you are calling set-up, I am calling translation. And to my mind translation follows after Identification. The identification and labelling step greatly simplifies step 2. When you just jump right into step 2, you have a much greater risk of error.

 

[mathdad]

Yes. So what you are calling set-up, I am calling translation. And to my mind translation follows after Identification. The identification and labelling step greatly simplifies step 2. When you just jump right into step 2, you have a much greater risk of error.

Ok. Thanks.